Variation of parameters: x^2y"+5xy'+3y=lnx y(1)=2 y'(1)=3

[craazyraiderz]

Ok I understand the basic concept of variation of parameters but im having a tough time finding my y1 and y2 for this problem

x^2y" + 5xy' + 3y = lnx y(1)=2 y'(1)=3

It says that I should guess y=x^r

Please help!

 

[Deleted member 4993]

Ok I understand the basic concept of variation of parameters but im having a tough time finding my y1 and y2 for this problem

x^2y" + 5xy' + 3y = lnx y(1)=2 y'(1)=3

It says that I should guess y=x^r

Please help!

After that hint - you cannot find y[sub:3cu8eaeu]1[/sub:3cu8eaeu] and y[sub:3cu8eaeu]2[/sub:3cu8eaeu]??!!

The homogeneous ODE is

x[sup:3cu8eaeu]2[/sup:3cu8eaeu]y" + 5xy' + 3y = 0

x[sup:3cu8eaeu]2[/sup:3cu8eaeu][r(r-1)x^[sup:3cu8eaeu]r-2[/sup:3cu8eaeu]] + 5x[rx[sup:3cu8eaeu]r-1[/sup:3cu8eaeu]] + 3x[sup:3cu8eaeu]r[/sup:3cu8eaeu]= 0

factor out x[sup:3cu8eaeu]r[/sup:3cu8eaeu] to get,

r[sup:3cu8eaeu]2[/sup:3cu8eaeu] + 4r + 3 = 0

(r+3)(r+1) = 0

Now continue.....

 

[galactus]

\(\displaystyle x^{2}y''+5xy'+3y=ln(x), \;\ y(1)=2, \;\ y'(1)=3\)

Playing on what SK said, divide by x^2

\(\displaystyle y''+\frac{5}{x}y'+\frac{3}{x^{2}}y=\frac{ln(x)}{x^{2}}\)

Let \(\displaystyle y=x^{m}\), then \(\displaystyle \frac{dy}{dx}=mx^{m-1}, \;\ \frac{d^{2}y}{dx^{2}}=m(m-1)x^{m-2}\)

\(\displaystyle m(m-1)x^{m-2}+5mx^{m-2}+3x^{m-2}=\frac{ln(x)}{x^{2}}\)

We can get the complementary function from \(\displaystyle m(m-1)+5m+3=(m+1)(m+3)\)

This gives \(\displaystyle m=-1, \;\ m=-3\)

Therefore, \(\displaystyle y_{c}=C_{1}x^{-1}+C_{2}x^{-3}\)

Now, continue with the V.O.P. noting that \(\displaystyle f(x)=\frac{ln(x)}{x^{2}}\)